If "NaCl" is dissolved in water to raise its boiling point by 2.10^@ "C", what is the molarity? K_b = 0.512^@ "C/m".

Jun 28, 2017

I assume you mean molality.

The boiling point elevation $\Delta {T}_{b}$ is defined in terms of the molality $m$, boiling point elevation constant ${K}_{b}$, and van't Hoff factor $i$:

$\Delta {T}_{b} = {T}_{b} - {T}_{b}^{\text{*}} = i {K}_{b} m$

The van't Hoff factor for $\text{NaCl}$ is ideally $2$ (the actual value is $1.9$), as 100% dissociation leads to two ions per formula unit, and thus the molality is:

DeltaT_b = 102.10^@ "C" - 100.00^@ "C" = (2)(0.512^@ "C"cdot"kg/mol")m

=>m = (2.10cancel(""^@ "C"))/(2 cdot 0.512cancel(""^@ "C")cdot"kg/mol")

$\implies \textcolor{b l u e}{m = \text{2.051 mol solute/kg solvent}}$

If you want the molarity, you'll have to provide the new volume of the solution.